Algebra 1: Common Core (15th Edition)

Published by Prentice Hall

Chapter 11 - Rational Expressions and Functions - 11-5 Solving Rational Equations - Practice and Problem-Solving Exercises - Page 695: 16

Answer

The equation has two solutions, 1 and 3.

Work Step by Step

$\frac{8}{x+3}=\frac{1}{x}+1$ $\frac{8x}{x(x+3)}=\frac{x+3}{x(x+3)}+1$ $\frac{8x}{x^2+3x}-\frac{x+3}{x^2+3x}=1$ $\frac{8x-x-3}{x^2+3x}=1$ $7x-3=x^2+3x$ $x^2-4x+3=0$ $x^2-3x-x+3=0$ $x(x-3)-(x-3)=0$ $(x-1)(x-3)=0$ $x-1=0$ or $x-3=0$ $x=1$ or $x=3$ Check: $8/(1+3)=8/4=2; 1/1+1=2$ $8/(3+3)=8/6=4/3; 1/3+1=4/3$ The equation has two solutions, 1 and 3.

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